Predicting the Direction of Reaction
N2(g) + 3H2(g)
2NH3(g) KP (472°C)=
(PNH3)2
(PN2)(PH2)3
Starting with 1.00 mol N2, 2.00 mol H2, and 2.00 mol NH3 in
a 1.00-L container at 472°C, which way will the reaction
proceed? KP(472°C) = 2.79 x 10-5.
Substituting these values into the equilibrium
expression gives a value known as the reaction
quotient Q.
𝑳 𝒃𝒂𝒓
R = 0.083145 𝒎𝒐𝒍 𝑲
(2.00 mol RT)2
Q=
1.00 L2
=
(1.00 mol RT)(2.00 mol RT)3
1.00 L
1.00 L3
124 2
62.0 1243
Q vs K
𝟐
N2(g) + 3H2(g)
2NH3(g)
𝑲𝑷 =
(𝑷𝑵𝑯𝟑 )𝒆𝒒
𝑷 𝑵𝟐
𝒆𝒒
𝟑
(𝑷𝑯𝟐 )𝒆𝒒
𝟐
𝑸=
(𝑷𝑵𝑯𝟑 )
𝑷𝑵𝟐
𝟑
(𝑷𝑯𝟐 )
• The expressions for K and Q look virtually identical.
 If K uses P (that is, if K is a KP), then Q uses P
(although Q is not given a subscript…be mindful!).
• The only allowed values in the expression for KP are
equilibrium values of P.
 KP is constant for a given T.
• Q’s value is not fixed. Any value of P may be used!
Predicting the Direction of Reaction
Using the Reaction Quotient Q
N2(g) + 3H2(g)
2NH3(g) KP (472°C)=
(PNH3)2
(PN2)(PH2)3
Starting with 1.00 mol N2, 2.00 mol H2, and 2.00 mol NH3 in a 1.00L container at 472°C, which way will the reaction proceed?
Q = 1.30 x 10-4
KP(472°C) = 2.79 x 10-5
Q > KP
This means the system contains a higher
concentration of products than would be present at
equilibrium. To reach equilibrium, some of the NH3,
must decompose into N2 and H2.
When Q > K, the system will react to use up
products and form reactants.
Predicting the Direction of Reaction
Using the Reaction Quotient Q
When Q > K, the system will react to use up
products and form reactants.
When Q < K, the system will react to use up
reactants and form products.
When Q = K, the system is at equilibrium and
the relative amounts of reactants and
products will not change.
Heterogeneous Equilibria
PbCl2(s)
K = [Pb2+] [Cl-]2
[PbCl2]
Pb2+(aq) + 2Cl-(aq)
Ksp = [Pb2+] [Cl-]2
What is the molarity of a pure substance? Or, more
accurately, what is the ratio of the concentration of
the pure substance to the standard value?
The ratio is constant and can be incorporated into K.
This applies when the pure substance is a solid or a
liquid.
Ksp is called the solubility product constant.
Heterogeneous Equilibria
PbCl2(s)
K = [Pb2+] [Cl-]2
[PbCl2]
Pb2+(aq) + 2Cl-(aq)
Ksp = [Pb2+] [Cl-]2
If a pure solid or liquid is involved in an equilibrium,
its concentration is not included on the right hand
side of the equilibrium expression.
Heterogeneous Equilibria
CaCO3(s)
CaO(s) + CO2(g)
What is the equilibrium
expression for this reaction?
KP = PCO2
Although CaCO3(s) and CaO(s) are not present in the
equilibrium expression for the reaction, they must be
present in the reaction system at equilibrium.
Heterogeneous Equilibria
CaCO3(s)
CaO(s) + CO2(g)
KP = PCO2
In any equilibrium, ALL substances must be present.
A SOLVENT IS CONSIDERED A PURE SUBSTANCE.
Heterogeneous Equilibria
CaCO3(s)
CaO(s) + CO2(g)
KP = PCO2
1. Starting with only CaCO3, could equilibrium
be reached?
Yes. CaO and CO2 will form as CaCO3
decomposes. However, there must be
enough CaCO3 to allow the equilibrium
PCO2 to be reached.
Heterogeneous Equilibria
CaCO3(s)
CaO(s) + CO2(g)
KP = PCO2
2. Starting with CaO and PCO2 > KP, could
equilibrium be reached?
Yes. PCO2 > KP means there is more than
the equilibrium amount of CO2 (Q > KP).
Therefore some CO2 will react with the CaO
to form CaCO3.
Heterogeneous Equilibria
CaCO3(s)
CaO(s) + CO2(g)
KP = PCO2
3. Starting with CaCO3 and PCO2 > KP, could
equilibrium be reached?
No. PCO2 > KP (Q > KP) means the
reaction favors the formation of
reactants and so no CaO will be formed.
4. Starting with CaCO3 and CaO, could
equilibrium be reached?
Yes.
Calculation of Equilibrium Concentrations
Calculate KC for the reaction
NH3(aq) + H2O (l)
NH4+(aq) + OH-(aq)
using the following information:
Enough ammonia is dissolved in 5.00 L of water at 25°C
to produce a solution that is 0.0124 M in ammonia. The
solution is then allowed to come to equilibrium. At that
time, the concentration of hydroxide ion is found to be
4.64 x 10-4 M.
KC = [NH4+] [OH-]
[NH3]
Calculating Equilibrium Concentrations
Enough ammonia is dissolved in 5.00 L of water at 25°C to produce a
solution that is 0.0124 M in ammonia. The solution is then allowed to
come to equilibrium. At that time, the concentration of hydroxide ion is
found to be 4.64 x 10-4 M. Calculate KC.
NH3(aq) + H2O (l)
Initial:
0.0124 M
NH4+(aq) +
0M
OH-(aq)
0M
Change: - 4.64 x 10-4 M
4.64 x 10-4 M
4.64 x 10-4M
Equilibrium: 0.011936 M
4.64 x 10-4 M
4.64 x 10-4M
KC = [NH4+] [OH-] = (4.64 x 10-4)2 = 1.80 x 10-5
[NH3]
0.011936
Calculating Equilibrium Concentrations
At 500°C an equilibrium mixture of gases in the Haber process was
found to contain a hydrogen partial pressure of 0.928 bar and a
nitrogen partial pressure of 0.432 bar. What is the partial pressure of
ammonia? KP(500°C) = 1.45 x 10-5
N2(g)
Equilibrium: 0.432 bar
+
3H2(g)
0.928 bar
2NH3(g)
x bar
KP = 1.45 x 10-5 = (PNH3)2
=
x2
=
x2
(PN2)(PH2)3 (0.432)(0.928)3
0.34525
x2 = (0.34525)(1.45 x 10-5) = 5.006 x 10-6
x = √(5.006 x 10-6) = 2.24 x 10-3 bar
Calculating Equilibrium Concentrations
A gas cylinder at 500 K is charged with phosphorus
pentachloride at an initial pressure of 1.66 bar. What
are the equilibrium pressures of PCl5, PCl3, and Cl2 at
this temperature? KP(500 K) = 0.497
PCl5(g)
PCl3(g)
Initial:
1.66 bar
0 bar
0 bar
Change:
-x bar
x bar
x bar
x bar
x bar
Equilibrium: 1.66 - x bar
KP = 0.497 =
(PPCl3)(PCl2)
(PPCl5)
=
+
x2
(1.66-x)
Cl2(g)
Calculating Equilibrium Concentrations
A gas cylinder at 500 K is charged with phosphorus
pentachloride at an initial pressure of 1.66 bar. What are the
equilibrium pressures of PCl5, PCl3, and Cl2 at this temperature?
KP(500 K) = 0.497
PCl5(g)
Equilibrium: 1.66 - x bar
KP = 0.497 =
(PPCl3)(PCl2)
(PPCl5)
PCl3(g)
x bar
=
+
Cl2(g)
x bar
x2
(1.66-x)
x2 = 0.497(1.66 – x)
x2 + 0.497x – 0.82502 = 0
x = 0.693, -1.19…which root is the correct one?
Hint: Look at the units for the answer.
Calculating Equilibrium Concentrations
A gas cylinder at 500 K is charged with phosphorus
pentachloride at an initial pressure of 1.66 bar. What
are the equilibrium pressures of PCl5, PCl3, and Cl2 at
this temperature? KP(500 K) = 0.497
PCl5(g)
Equilibrium 0.97 bar
PCl3(g)
0.693 bar
+
Cl2(g)
0.693 bar
Total pressure at equilibrium
Ptot = 0.97 + 0.693 + 0.693 = 2.35 bar
Le Châtelier’s Principle
The conclusions we draw from the size of the reaction
quotient Q relative to the equilibrium constant K tell us
which way the system will move to reach equilibrium. We
are, in effect, using Le Châtelier’s Principle.
Le Châtelier’s Principle
If a system at equilibrium is disturbed by a
change in temperature, pressure, or the
concentration of one of the components, the
system will shift its equilibrium position so as
to counteract the effect of the disturbance.
Le Châtelier’s Principle
Changing reactant or product concentrations
N2(g) + 3H2(g)
2NH3(g)
KP (472°C)=
(PNH3)2
(PN2)(PH2)3
How would adding nitrogen to the equilibrium mixture
change the partial pressures of the gases?
At equilibrium, Q = KP . Adding N2 would cause Q < KP .
The equilibrium would shift to the right, using up some of
the N2 and H2 and creating more NH3 . So the partial
pressure of H2 would decrease and the partial pressure of
NH3 would increase. We would have to calculate the effect
on the partial pressure of N2 .
Le Châtelier’s Principle
Changing reactant or product concentrations
N2(g) + 3H2(g)
2NH3(g)
KP (472°C)=
(PNH3)2
(PN2)(PH2)3
How would removing ammonia from the equilibrium
mixture change the partial pressures of the gases?
At equilibrium, Q = KP . Removing NH3 would cause Q
< KP . The equilibrium would shift to the right, using
up some of the N2 and H2 and creating more NH3.
Removing product, in this case, is a way to generate
more product.
Le Châtelier’s Principle
Changing reactant or product concentrations
2H+(aq) + 2CrO42- (aq)
Cr2O72-(aq) + H2O (l)
How would adding HCl(aq) affect this equilibrium?
How would adding NaOH(aq) affect this equilibrium?
Le Châtelier’s Principle
Decreasing the volume of a system (this has
the effect of increasing the pressure).
N2(g) + 3H2(g)
2NH3(g)
KP (472°C)=
(PNH3)2
(PN2)(PH2)3
The reaction will shift toward the side with
fewer moles of gas.
Le Châtelier’s Principle
Decreasing the volume of a system (this has
the effect of increasing the pressure).
N2(g) + 3H2(g)
2NH3(g)
KP (472°C)=
(PNH3)2
(PN2)(PH2)3
In this reaction, the product side has fewer moles of
gas. So decreasing the volume will increase PNH3 .
OR At equilibrium, Q = KP . If the volume were halved,
the partial pressures would all double, resulting in Q <
KP . The equilibrium would shift to the right.
Le Châtelier’s Principle
Adding an inert gas to a reaction system
Since an inert gas does not participate in the
reaction, the equilibrium will not change. The
only thing that changes is the total pressure.
N2(g) + 3H2(g)
2NH3(g)
KP (472°C)=
(PNH3)2
(PN2)(PH2)3
Le Châtelier’s Principle
Changing the temperature of a reaction system
This is a change that affects the value of K
itself.
Increasing T effectively adds heat to the system.
PCl5(g)
PCl3(g) + Cl2(g)
PCl5(g) + 87.9 kJ
ΔH° = 87.9 kJ
PCl3(g) + Cl2(g)
If the reaction is endothermic, increasing T will
increase K, shifting the equilibrium to the right.
Le Châtelier’s Principle
Changing the temperature of a reaction system
This is a change that affects the value of K
itself.
Look up in Appendix C
ΔH°(25°C) = ? kJ
N2(g) + 3H2(g)
2NH3(g)
N2(g) + 3H2(g)
2NH3(g) + 92.38 kJ
If the reaction is exothermic, increasing T will
decrease K, shifting the equilibrium to the left.
Le Châtelier’s Principle
The effect of a catalyst
Catalysts change reaction rates, not
equilibrium mixtures. The use of a
catalyst will not change K or the
equilibrium mixture.
Integrating Concepts
C(s) + H2O(g)
CO(g) + H2(g) KP(800°C) = 14.1
1. If we start with carbon and and 0.100 mol water vapor in a
1.00-L vessel, what will the partial pressures of the gases
be at equilibrium?
2. What is the minimum amount of carbon needed to achieve
equilibrium at 800°C?
3. What will the total pressure inside the vessel be at
equilibrium?
4. At 25°C, KP for the reaction is 1.7 x 10-21. Is the reaction
endothermic or exothermic?
5. To maximize yield at 800°C, should the volume of the
system be increased or decreased?
Integrating Concepts
1. If we start with carbon and and 0.100 mol water vapor in a 1.00-L
vessel, what will the partial pressures of the gases be at
equilibrium?
C(s) +
H2O(g)
CO(g) +
0.100 mol
-x mol
0 mol
x mol
0 mol
x mol
0.100-x mol
x mol
x mol
Initial: present
Change: -x mol
Equilibrium: ?
H2(g)
KP (800°C) = 14.1 = (PCO)(PH2) = (xRT/V)2
= x2RT
(PH2O)
(0.100-x)RT/V (0.100-x)V
1.41 - 14.1x = x2RT/V
RT/V = 89.227 (how ??)
89.227x2 + 14.1x - 1.41 = 0
Integrating Concepts
1. If we start with carbon and and 0.100 mol water vapor in a 1.00-L
vessel, what will the partial pressures of the gases be at
equilibrium?
C(s) +
H2O(g)
CO(g) +
0.100 mol
-x mol
0 mol
x mol
0 mol
x mol
0.100-x mol
x mol
x mol
Initial: present
Change: -x mol
Equilibrium: ?
x 
 14 . 1 
14 . 1  4 ( 89 . 227 )(  1 . 41 )
H2(g)
2
2 ( 89 . 227 )
P(steam) = 2.72 bar
x = 0.0695 mol, -0.227 mol
P(CO) = P(H2) = 6.20 bar
Integrating Concepts
2. What is the minimum amount of carbon needed to achieve
equilibrium at 800°C?
C(s) +
H2O(g)
Equilibrium: present 0.0303 mol
0.0695 mol CO x
CO(g)
+
H2(g)
0.0697 mol 0.0697 mol
1 mol C x 12.01 g C = 0.835 g
1 mol CO
1 mol C
There must be a little more than 0.835 g C present at the
start of the reaction.
Integrating Concepts
3. What will the total pressure inside the vessel be at equilibrium?
C(s) +
Equilibrium: 0 bar
H2O(g)
CO(g)
2.72 bar
6.20 bar
+
Ptot = 2.72 + 6.20 + 6.20 = 15.12 bar
H2(g)
6.20 bar
Integrating Concepts
C(s) + H2O(g)
CO(g) + H2(g) KP(800°C) = 14.1
4. At 25°C, KP for the reaction is 1.7 x 10-21. Is the reaction
endothermic or exothermic?
KP increases with increasing T, so
the reaction is endothermic.
Integrating Concepts
C(s) + H2O(g)
CO(g) + H2(g) KP(800°C) = 14.1
5. To maximize yield at 800°C, should the volume of the
system be increased or decreased?
Increasing the volume favors formation of products
(because it decreases the partial pressures of all of the
gases and causes Q < KP. The reaction shifts to the
right.)